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Is it possible to choose an infinite GP from amongst the terms 1, 1/2, 1/4, 1/8, 1/16 ... with a sum a) 1/5? b) 1/7 ?

My approach was simply choosing terms and probable ratios to match the sum...I would be grateful if you could reveal any trick for this question?

Tanishka Marrott
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3 Answers3

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Since all terms are of the form $\frac{1}{2^n}$, any geometric progression will be with ratio $\frac{1}{2^k}$

The summation of a geometric series is:

$$\sum\limits_{k=0}^\infty ar^k = \frac{a}{1-r}$$

The question then is what choice of $a$ (initial term) and what choice of $r$ (ratio) will give the desired results.

Since $a$ and $r$ will both be of the form $\frac{1}{2^n}$ we have the sum must be of the form

$$\frac{\frac{1}{2^m}}{1-\frac{1}{2^n}} =\frac{2^{n-m}}{2^n-1}$$

for some given $n$ and $m$

The question is, can you find non-negative integers $n$ and $m$ such that $\frac{2^{n-m}}{2^n-1}$ is equal to $\frac{1}{5}$? is equal to $\frac{1}{7}$? If not, then why? If yes, which?

For $\frac{1}{5}$, the question becomes whether we can find $a,b$ such that $5\cdot 2^a = 2^b-1$. Notice that for $a\geq 1$ and $b\geq 1$ the left side is even while the right side is odd. If $b=0$ the right side is zero and the left side is positive. If $a=0$, the left side is equal to $5$, but no integer $b$ satisfies $2^b-1=5$

Thus $\frac{1}{5}$ is impossible.

For $\frac{1}{7}$, we ask if we can find $a,b$ such that $7\cdot 2^a=2^b-1$. From the same logic, we see that they cannot both be positive and $b$ cannot be zero. This leaves $a=0$ and we find that $b=3$ works.

This corresponds back to our $n$ and $m$ from before implying the geometric progression $\frac{1}{8},\frac{1}{64},\frac{1}{512},\dots$ has sum equal to $\frac{1}{7}$

Community
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JMoravitz
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Nothe that sum of an infinite G.P. is given by $\displaystyle\sum_{n=1}^{\infty} ar^{n-1}=\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

Note that in the series $\displaystyle\frac{1}{1},\frac{1}{2},\frac{1}{4}...$ all terms have a common ratio of the form $\frac{1}{2k}$ where $k\in\mathbb Z^+$.Or,all possible common ratios are $\frac{1}{2}$ or $\frac{1}{4}$ or $\frac{1}{8}$ and so on.This is because whenever we divide any two terms in the series we get a number of the form $\frac{1}{2k}$.

So,$$\frac{a}{1-\frac{1}{2k}}=\frac{1}{5}$$

$$\implies a=\frac{2k-1}{10k}$$

Now,$\frac17$ or $\frac15$ has $1$ in numerator.

So,$k=1$ in $a$ for numerator to become $1$.Plugging $k=1$,we get $a=\frac{1}{10}$.Now,this is not a term of the series.So,we have disproved one case.

Do the same for the other case.

Hope this helps!!

Soham
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A geometric series picked from the given series will have initial term $\frac1{2^a}$ ($a\ge0$) and ratio $\frac1{2^r}$ ($r\ge1$). The sum of this sub-series will be $$\frac{\frac1{2^a}}{1-\frac1{2^r}}=\frac{2^{r-a}}{2^r-1}$$ We cannot pick a geometric series for (a) that sums to $\frac15$ because a power of two (the numerator) multiplied by 5 cannot equal a Mersenne number (the denominator); this may be seen by comparing binary expansions.

For (b), however, we can find a geometric series summing to $\frac17$; take $r=a=3$.

Parcly Taxel
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